I want to maximize $$\Phi_g(w):=\sum_{i\in I}\sum_{j\in I}\int\lambda({\rm d}x)\int\lambda({\rm d}y)\left(w_i(x)p(x)q_j(y)\wedge w_j(y)p(y)q_i(x)\right)\sigma_{ij}(x,y)|g(x)-g(y)|^2$$ over all choices of $w=\left(w_1,\ldots,w_{|I|}\right)$ subject to $$\sum_{i\in I}w_i=1\tag1.$$ I guess this is usually solved by the method of Lagrange multipliers, but the shape of the integrand seems to be problematic. What can we do?

If this problem is too hard, are we at least able to find a choice of $w$ yielding a sharp lower bound?

As usual, $a\wedge b:=\min(a,b)$. And it might be useful to rewrite $\Phi_g(w)$ using $2(a\wedge b)=a+b-|a-b|$.

The objects are defined as follows:

• $(E,\mathcal E,\lambda)$ is a measure space
• $I$ is a finite nonempty set
• $p,q_i:E\to[0,\infty)$ are $\mathcal E$-measurable with $$\int p\:{\rm d}\lambda=\int q_i\:{\rm d}\lambda=1\tag2$$ for $i\in I$
• $g\in L^2(p\lambda)$
• $w_i:E\to[0,1]$ is $\mathcal E$-measurable with $$\{q_i=0\}\subseteq\{w_ip=0\}\tag3$$ for $i\in I$ with $$\{p\ne0\}\subseteq\left\{\sum_{i\in I}w_i=1\right\}\tag4$$
• $\sigma_{ij}:E^2\to[0,\infty)$ is $\mathcal E^{\otimes2}$-measurable for $i,j\in I$ with $$\sigma_{ij}(x,y)=\sigma_{ji}(y,x)\;\;\;\text{for all }x,y\in E\text{ and }i,j\in I\tag5$$ and $$\sum_{j\in I}\int\lambda({\rm d}y)q_j(y)\sigma_{ij}(x,y)=1\tag6$$

I'm primarily interested in finding the choice of $w=\left(w_1,\ldots,w_{|I|}\right)$ maximizing $\Phi_g(w)$ and satisfying $(3)$ and $(4)$, but if it's easier to deal with feel free to assume $(1)$ instead of $(4)$.